Question

# Two identical metal plates are given positive charge $${Q}_{1}$$ and $${Q}_{2}$$ $$\left( <{ Q }_{ 1 } \right)$$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $$C$$, the potential difference between them is

A
(Q1+Q2)/2C
B
(Q1+Q2)/C
C
(Q1Q2)/C
D
(Q1Q2)/(2C)

Solution

## The correct option is D $$\left( { Q }_{ 1 }-{ Q }_{ 2 } \right) /(2C)$$The potential difference between the two identical metal plates is given as $$C = \dfrac{{{\varepsilon _{\rm{0}}}A}}{d}$$ Let the surface charge density is given as $${\sigma _1} = {\sigma _{21}} = \dfrac{Q}{A}$$ The net electric field is $${E_{net}} = \dfrac{{{\sigma _1} - {\sigma _2}}}{{2{\varepsilon _{\rm{0}}}}}$$ We know the potential difference is given as $$V = E.d$$ By substituting the above values we get $$V = \dfrac{{{Q_1} - {Q_2}}}{{2C}}$$  Physics

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